03-树3 Tree Traversals Again（25 分）

An inorder binary tree traversal can be implemented in a non-recursive way with a stack. For example, suppose that when a 6-node binary tree (with the keys numbered from 1 to 6) is traversed, the stack operations are: push(1); push(2); push(3); pop(); pop(); push(4); pop(); pop(); push(5); push(6); pop(); pop(). Then a unique binary tree (shown in Figure 1) can be generated from this sequence of operations. Your task is to give the postorder traversal sequence of this tree.

Figure 1

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (≤30) which is the total number of nodes in a tree (and hence the nodes are numbered from 1 to N). Then 2N lines follow, each describes a stack operation in the format: “Push X” where X is the index of the node being pushed onto the stack; or “Pop” meaning to pop one node from the stack.

Output Specification:

For each test case, print the postorder traversal sequence of the corresponding tree in one line. A solution is guaranteed to exist. All the numbers must be separated by exactly one space, and there must be no extra space at the end of the line.

Sample Input:

6
Push 1
Push 2
Push 3
Pop
Pop
Push 4
Pop
Pop
Push 5
Push 6
Pop
Pop

Sample Output:

3 4 2 6 5 1

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注意的地方：

三种非递归 push pop 先序 先序 中序 中序 先序 中序 后序 先序 后序

push和pop的数字序列，后序是使用标记变量来实现的。

所以题目中push就是先序，pop就是中序，因此可以唯一确定后序。

依照递归来实现，ipush用来确定root，iin确定新的左子树个数，ipost确定偏移量（到底是哪一块的最后一个）。

#include <iostream>#include <stack>#define MAXSIZE 30//#define LOCALint push[MAXSIZE];int inOrder[MAXSIZE];int post[MAXSIZE];void fuc(int ipush, int iin, int ipost, int num);int main(){#ifdef LOCAL    freopen("test.txt", "r", stdin);#endif    int N;    char input[5];    std::stack<int> s;    std::cin >> N;    int index_push = 0, index_in = 0;    for(int i=0; i<2*N; ++i) {        scanf("%s", input);        int data;        if(input[1]=='u') {            std::cin >> data;            push[index_push++]= data;            s.push(data);        }        else {            inOrder[index_in++] = s.top();            s.pop();        }    }    fuc(0,0,0,N);    for(int i=0; i<N; ++i) {        if(i) std::cout << ' ';        std::cout << post[i];    }}void fuc(int ipush, int iin, int ipost, int num) {    if(!num) return;    if(num == 1) {        post[ipost] = inOrder[iin];        return;    }    int root = push[ipush];    //ipost是offset    post[ipost+num-1] = root;    int i, L, R;    for(i=0; i<num; ++i) {        if(inOrder[iin + i]==root) break;    }    L = i; //左子树个数    R = num - i - 1; //右子树个数    fuc(ipush + 1, iin, ipost, L);    //         前   左      左   前   偏移量更新L    fuc(ipush + 1 + L, iin + L + 1, ipost + L, R);}